Exam CV0-003 topic 1 question 101 discussion

To meet the minimum IP requirement for the described deployment scenario, you need to account for a total of 7 IP addresses (2 web servers + 2 database servers + 2 additional web servers + 1 load balancer). Additionally, you need to include the gateway and network/broadcast addresses, which means you'll need at least 11 IP addresses in total. The best option that meets these requirements is: B. 192.168.1.0/27 A /27 subnet provides 32 IP addresses (30 usable addresses after accounting for the network and broadcast addresses), which is more than sufficient for the deployment scenario and future expansion.

Some of you guys are going to fail your Net+ or anything else network related. Here's the proper breakout. C is indeed correct, but some of these comments are just terrible and no one has laid it out, so here it is: 1 - gateway (this will be our public IP but will have internal as well as all gateways always do!) 2 - load balancer 3 - web server #1 4 - web server #2 5 - database server 6 - database server 7 - new webserver #1 8 - new web server #2 9 - broadcast IP Our /28 is going to provide a maximum of 16 IPs which will cover the 9 addresses needed for current and future expansion.

The load balancer will have a public IP. Doesn't that exclude it from the private network, leading to /29 being sufficient for 6 hosts

7 addresses are needed, /28 provides 8 addresses but only 6 of those are available after discounting the network and broadcast addresses. So /29 is the smallest subnet that meets requirements.

28 is correct

Total IP requirement is 7 including future growth so a /29 can provide 8 IPs. So why a /28?